Question #ec1c9

#2sin(5x)=-sqrt2#

Lets isolate our #sine#:

#2sin(5x)=-sqrt2#

divide by #2# on both sides
#sin(5x)=-sqrt(2)/2#

So, we are looking for a number that, when multiplied by #5#, will give us the ratio of #-sqrt(2)/2#. That sounds like a lot, but let's break it down.

First, the value #-sqrt(2)/2# corresponds to #45^0# or #pi/4#.

I like to use degrees :)
So, we need to find a number that, multiplied by #5#, equals #45#.

#5*9=45#

#x=9#

#sin(5*9)=-sqrt(2)/2#
#sin(45)=-sqrt(2)/2#

EDIT: I do believe that it is a mistake to use degrees (you can but you need to inlcude more work from the pi/180 conversion factor). Basically the answer should be in radians.

So, we're looking for a number that, when multiplied by 5, equals #-sqrt(2)/2# when plugged into sin(x).

So, when does #sin(x)=-sqrt(2)/2#?
Well, sin(x) is negative in the 3rd and 4rth Quadrants. (This is something you kinda have to memorize, though there are patterns that will help you.)

Next, #sqrt(2)/2#, from any #Sin(x)# or #Cos(x)# is #pi/4#

Finally, #pi/4# in the 3rd and 4rth quadrants are:
3rd: #pi+pi/4= 5pi/4#
4rth: #2pi-pi/4= 7pi/4#

so #x=5pi/4+-2pi#
and #x=7pi/4+-2pi#

we have "#+-2pi#" at the end because every #2pi# is the cycle around the unit circle, resulting in the same answer.